Sin(2θ)+sin(<span>θ)=0
use double angle formula: sin(2</span>θ)=2sin(θ)cos(<span>θ).
=>
2sin(</span>θ)cos(θ)+sin(<span>θ)=0
factor out sin(</span><span>θ)
sin(</span>θ)(2cos(<span>θ)+1)=0
by the zero product property,
sin(</span>θ)=0 ...........(a) or
(2cos(<span>θ)+1)=0.....(b)
Solution to (a): </span>θ=k(π<span>)
solution to (b): </span>θ=(2k+1)(π)+/-(π<span>)/3
for k=integer
For [0,2</span>π<span>), this translates to:
{0, 2</span>π/3,π,4π/3}
You probably jsut need to use law of cosines in order to solve for angle A
or
Answer:
1/2
Step-by-step explanation:
(1/16)^x
Let x = 1/4
(1/16)^ 1/4
Rewriting 16 as 2^4
(1/2^4)^ 1/4
We know that 1 / a^b = a^-b
(2 ^ -4)^ 1/4
We know that a^b^c = a^(b*c)
2^(-4*1/4)
2^-1
We know that a^-b = 1/ a^b
2^-1 = 1/2^1 = 1/2
1.3 pages first you take 15 * by 60 because 60 min. in one hour and then divide 1200 by 900 you get 1.3
